The position of the centre of mass of a uniform semi-circular wire of radius 'R' placed in x-y plane with its centre at the origin and the line joining its ends as x-axis is given by ( 0,xR/π ). Then, the value of | x | is ______________.

Solution

Physics sir · 4 steps

Cut the arc into tiny pieces, measure each by its angle, and add up their heights.

1. Set up the arc. Measure the angle θ from the positive x-axis, so the wire runs from θ = 0 to θ = π. A tiny piece at angle θ has length R dθ and mass dm = λ R dθ, where λ = M/(π R). Its height is y = R sin θ.
2. Symmetry first. The wire is symmetric about the y-axis, so x_(cm) = 0. Only y_(cm) needs working out.

   y_(cm) = 1/M∫₀^π (R sin θ) λ R dθ = (λ R²)/M[-cos θ]₀^π = (2λ R²)/M

3. Put in lambda. With λ = M/(π R), this gives y_(cm) = 2R²/M × M/(π R) = 2R/π.
4. So. The centre of mass is at (0, 2R/π). Matching this with (0, xR/π) gives |x| = 2.

The common slip: 1.33. Using the result for a semicircular disc, 4R/3π, instead of the one for a wire gives 4/3 ≈ 1.33, not 2.

The idea to keep. Learn these two side by side: a semicircular ring (wire) has its centre of mass at 2R/π, and a semicircular disc (lamina) has it at 4R/3π. Check whether the question means a wire or a sheet.

Answer from the NTA final answer key. Solution by PrepWiser.